Giva an example of correspondence $F : \mathbb{R} \rightarrow \mathbb{R}$ such that the closure of $F$ is $ \overline{F}: \mathbb{R} \rightarrow \mathbb{R}$, upper semi continuous on $\mathbb{R}$, but $F$ is not, if any.
I got stuck with this question. Any help will be appreciated.
$\def\mto{\multimap}\def\R{\mathbb R}$Define $F \colon \R \mto \R$ by $$ F(x) = \begin{cases} [0, 1] & x > 0\\ [0,1) & x \le 0\end{cases} $$ Then $F$ isn't upper semi-continuous, as $(-1,1)$ is an open neighbourhood of $F(0)$, but for each open neighbourhood $U$ of $0$ we have $F(U) \not\subseteq (-1,1)$. The closure $$ \bar F(x) = [0,1], \quad x \in \mathbb R $$ is upper semi-continuous, as it is constant.
